Answer:
They are called corpuscles
Explanation:
Answer:
True.
Explanation:
To know which option is correct, let us calculate the number of mole present in 60g of calcium. This is illustrated below:
Mass of Ca = 60g
Molar Mass of Ca = 40g/mol
Number of mole Ca =....?
Number of mole = Mass/Molar Mass
Number of mole of Ca = 60/40
Number of mole Ca = 1.5 moles.
From the calculations made above, we can see that 1.5 moles are present in 60.0 grams of calcium
Answer:
1. Ionic bonding
2. Covalent bonding
3. Metallic bonding
Explanation:
Ionic bonding also referred to as electrovalent bonding is a kind of chemical bonding that involves the transfer of electrons between the valence shells of two elements with a large electronegativity difference usually a metal and a nonmetal.
For example an ionic bonding scenario might play out between a group one metal and a group seven halogen. While group one metals have one electron hindering their stability, group seven halogens need that one electron that could make them achieve this stability. It is this that causes them to come together in a way where the electron is transferred completely from the valence shell of the group 1 atom and accepted into the valence shell of the group 7 halogen.
Covalent bonding involves the sharing of electrons between atoms of comparable electronegativities. The electro negativity difference is not large enough to permit the total movement of the electrons and hence the electrons are then controlled by the nuclei of the two atoms
Between two metals, what we have is called the metallic bonding
Answer:
'See Explanation
Explanation:
Determine the [OH−] , pH, and pOH of a solution with a [H+] of 9.5×10−13 M at 25 °C.
Given [H⁺] = 9.5 x 10⁻¹³M => [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ => [OH⁻] = 1.0 x 10⁻¹⁴/9.5 x 10⁻¹³ = 0.0105M
pH = -log[H⁺] = -log(9.5 x 10⁻¹³) = - (-1202) = 12.02.
pOH = -log[OH⁻] = -log(0.0105) = -(-1.98) = 1.98
Now you use the same sequence in the remaining problems.
Answer:
The answer is C. An electrochemical cell.
Explanation:
The aluminum ion react with the sulfide to form aluminum sulfide.
